Question

The escape velocity on a planet with radius double that of earth and mean density equal to that of earth will be (escape velocity on earth $ = 11.2Km/s $ )
(A) $ 11Km/s $
(B) $ 22Km/s $
(C) $ 5.5Km/s $
(D) $ 15.5Km/s $

Answer 1

Hint
The escape velocity is given by the formula, $ v = \sqrt {\dfrac{{2GM}}{R}} $ . So for the earth, we can substitute the mass of earth with an expression in terms of the density and the radius of the earth. Then we can find the escape velocity of the other planet in terms of the escape velocity of the earth by using the same formula and hence calculate the escape velocity on that planet.
In this solution, we will be using the following formula,
$\Rightarrow v = \sqrt {\dfrac{{2GM}}{R}} $
where $ v $ is the escape velocity, $ G $ is the universal gravitational constant, $ M $ is the mass and $ R $ is the radius of any planet.
$\Rightarrow M = \dfrac{4}{3}\pi {R^3}\rho $ where $ \rho $ is the mean density of that planet.

Complete step by step answer
The escape velocity of the earth is given by the formula
$\Rightarrow v = \sqrt {\dfrac{{2GM}}{R}} $
Now, the mass of the earth is dependent on the density of the earth and the radius of the earth. It is given by the following formula,
$\Rightarrow M = \dfrac{4}{3}\pi {R^3}\rho $
So by substituting the value of $ M $ in the formula for the escape velocity, we get
$\Rightarrow v = \sqrt {\dfrac{{2G\dfrac{4}{3}\pi {R^3}\rho }}{R}} $
We can cancel the $ R $ from the numerator and the denominator, so we get
$\Rightarrow v = \sqrt {\dfrac{{8G\pi {R^2}\rho }}{3}} $
The $ {R^2} $ can come out of the root as $ R $ . Hence, we get
$\Rightarrow v = R\sqrt {\dfrac{{8G\pi \rho }}{3}} $
This is the escape velocity of the earth.
Now we consider a planet whose mean density is equal to that of the earth and the radius is twice of earth. So, $ R' = 2R $ and $ \rho ' = \rho $ .
Therefore from the formula of escape velocity, we get the escape velocity of this planet as,
$\Rightarrow v' = R'\sqrt {\dfrac{{8G\pi \rho '}}{3}} $
Substituting the values we get,
$\Rightarrow v' = 2R\sqrt {\dfrac{{8G\pi \rho }}{3}} $
Therefore in place of $ R\sqrt {\dfrac{{8G\pi \rho }}{3}} $ we can write $ v $ .
So, $ v' = 2v $
Now the escape velocity of earth is given as, $ v = 11.2km/s $
Therefore,
$\Rightarrow v' = 2 \times 11.2km/s $
hence, we get the escape velocity of the planet as,
$\Rightarrow v' = 22.4km/s $ which is approximately equal to $ v' = 22km/s $
So the correct option is (B); $ 22km/s $

Note
The escape velocity on the surface of any planet is the minimum speed that is required by any object to escape the gravitational influence of that planet. The escape velocity depends on the mass of the planet but not the mass of the object that is trying to escape the gravitational influence of that planet.